I was wondering, what a $N$-dimensional simplicial space $X$ should be. Of course the degeneracy maps force the spaces to be nonempty in high dimensions. Currently I have two different versions and i am wondering whether they are equivalent:

1) There are no nondegenerate simplexes above dimension $N$.

2) Let $\Delta|_N$ be the full subcategory of $\Delta$ consisting of the objects $[0],\ldots ,[N]$. The inclusion induces a restriction functor $R$ from $\Delta-$spaces to $\Delta|_N$-spaces, which has a left adjoint $L$. There is a canonical map $L(R(X))\rightarrow X$. $X$ should be called $N$-dimensional, iff this map is an isomorphism,i.e. a homeomorphism on each object of $\Delta$.

To make the question more precise: $2)\Rightarrow 1)$ can be easily seen, as the map $L(R(X))\rightarrow X$ cannot hit any nondegenerate simplex above dimension $N$ by construction. The other way round:
Given $1)$, then all the maps occuring in the natural transformation are surjective and continuous. Injectivity should follow from the relations in $\Delta$. So why is the inverse map of sets continuous?

(In the category of simplicial sets (and not spaces) both notions should be equivalent using the same argumentation.)

$\begingroup$Is your map $X\to LR(X)$ going the right way? If L is left adjoint then surely the map goes from $LR(X)$ to $X$ (I think: 'Free group on underlying set goes to group'!) The map is then, geometrically, the inclusion of the N-skeleton. That I suspect from later on is a typo. I think you will find the answer if you describe L explicitly.$\endgroup$
– Tim PorterFeb 12 '10 at 18:41

$\begingroup$oops sry. you are right .$\endgroup$
– HenrikRüpingFeb 12 '10 at 18:49

3 Answers
3

I'm cautiously optimistic that 1) $\Rightarrow$ 2). As you say, if $X$ satisfies 1), then $L(R(X))\to X$ is a continuous bijection. (Because it's a statement about point-sets, and the thing is true for simplicial sets.)

If $N=0$, it should be easy: $L(R(X))$ is the constant simplicial space, whose value at each [k] is $X_0=X([0])$. The canonical map $L(R(X))\to X$ is the one which at degree $k$ is given by the map $s:X_0\to X_k$ defined by the composite of degeneracy operators. But we know that the composite $X_0\to X_k\to X_0$ is the identity, where $d:X_k\to X_0$ is a composite of face operators. So if $s$ is a bijection, $d$ is its continuous inverse.

For general $N$, let $Y=L(R(X))$. The functor $R$ also has a right adjoint, which I'll call $M$. Let $Z=M(R(X))$. Just as the space $Y_k$ looks like a colimit of a certain diagram of the spaces $X_0,\dots,X_N$, the space $Z_k$ looks like a limit of a certain diagram of these spaces.

There are canonical maps $Y=L(R(X)) \to X \to M(R(X))=Z$. I would like to claim that the composite $f:Y\to Z$ gives a homeomorphism of $Y$ onto its image. If you can prove this, that will give your result, since a continuous inverse to $Y\to X$ will be given by $X\to f(Y)\approx Y$.

$\begingroup$It may help for comparison with known results to us the skeleton and coskelton functor terminology here. $\endgroup$
– Tim PorterFeb 12 '10 at 21:54

1

$\begingroup$Good point. Specifically, L(R(X)) is the Nth "skeleton" of X, and M(R(X)) is the Nth "coskeleton" of X.$\endgroup$
– Charles RezkFeb 13 '10 at 0:08

1

$\begingroup$@ Henrik As 'dimension' could be ambiguous, perhaps some sort of name such as N-skeletal might be better for this concept. My worry is that if the spaces of simplices have themselves a dimension there is a risk of confusion. As a 'for instance', if we start with a bisimplicial set and realise geometrically in one direction, then the question still makes sense, yet in that `realised' direction no restriction is being made.$\endgroup$
– Tim PorterFeb 13 '10 at 10:09

$\begingroup$i agree. The dimension of a simplicial space or set really depends on the chosen functor and not only on the geometric realisation. But i have some hope, that for good spaces of simplices the dimension agrees with some dimension of the realisation. Currently I believe, that if the space of simplices is totally disconnected Hausdorff, then the "dimension" of the simplicial space agrees with the topological dimension of its realisation.$\endgroup$
– HenrikRüpingFeb 13 '10 at 21:51

1

$\begingroup$There are results on profinite simplicial spaces, but not ones relating to dimension of a realisation. I have looked at profiniteness from various angles but generally it seemed not that good an idea (for what I wanted to do) to take geometric realisations. You may have a good reason to want to do that but it does seem a strange thing to me. The category of profinite simplicial spaces is very nice for doing lots of things, and I never had the desire to form some geometric realisation of one of them.$\endgroup$
– Tim PorterFeb 15 '10 at 17:01

OK I checked, how the adjoint functors looks like. Given any $\Delta|_N $ simplicial space $X$. To define $L(X)$, we have to extend $X$ to the whole category $\Delta$. I am just telling, what $L(X)$ does on $[N+1]$. Then you keep extending the functor in the same way:

$L(X)([N+1]):=(0,\ldots,N)\times X([N])/\sim$, where the equivalence relation is given by
$(j,s_k(x))\sim (k+1,s_j(x))$ for $0\le j\le k\le N,x\in X[N-1]$. The $i$-th degeneracy map is induced by the inclusion of the i-th summand. Using the relations in $\Delta$ one can also define the face maps.

The right adjoint functor is given by
$M(X)([N+1]):= ( (x_0,\ldots,x_{N+1})|\partial_ix_j=\partial_{j-1}x_i\mbox{ for } 0 \le i < j \le N+1 )\subset \prod_{i=0}^{N+1}X[N]$. The face maps are just the projections and one can define the degeneracy maps using the relations in $\Delta$.

So let $X$ be a $\Delta$-space. The natural transformation is given by
$L(R(X))([N+1])\rightarrow M(R(X))([N+1])\qquad (i,x)\mapsto (\partial_0 s_i(x),\ldots,\partial_{N+1} s_i(x))$.

Using the relations in $\Delta$ one can show, that this map is injective. So the remaining question is, whether this map is an open map (considered as a map onto the image).

I just wanted to write a comment, that I could show the upper equivalence conly for simplicial sets and simplicial Hausdorff spaces (but this answer is too long for a comment).

This question still makes sense in other categories replacing 1) by

1') The span of all degenerate subobjects $s_0(X([N-1])),\ldots,s_{N-1}(X([N-1]))$ is the whole of $X([N-1])$.

For example it fails in the category of simplicial groups. There are two simplicial groups $X$ and $Y$ , whose restrictions to $\Delta|_1$ agree. But if you consider the span of all degenerate elements in $X([2])$ resp. $Y([2])$ those spans are not isomorphic.
Consider the simplicial set $S$ with one 0-simplex $a$ and 1 nondegenerate 1-simplex $b$. Define $X([k]):=Y([k]):=\mathbb{Z}(S([k])$ for $k=0,1$. Then you can define $X([2])$ as $\mathbb{Z}(a,s_0(b),s_1(b))$ and $Y([2])$ as $\mathbb{Z}(a)\times F(s_0(b),s_1(b))$, where $F(s_0(b),s_1(b))$ denotes the free group in $s_0(b),s_1(b)$.

The face and degeneracy maps in $S$ induce in both cases well defined face and degeneracy maps. Hence this shows, that $X$ and $Y$ give a counterexample.